Surface Area Integral

Find the area of the surface z=7+2x+2y2 that lies above the triangle with vertices (0,0) (0,1) and (2,1)

3. The attempt at a solution

It's not a particularly difficult problem to set up, I just can't seem to get a simple integrand.

∫∫√(16y2+5) dA is what I come up with, and I'm not quite sure how to find the anti-derivative of that. dA=dy dx, 0<y<(x/2) & 0<x<2.

I can't do a u-sub because I don't have anything to cancel out dy with. I've looked in some integral tables also, and the closest thing I could think of would give me an arcsin function and I'm pretty sure that's not how I'm supposed to do it.

First integrate with respect to x, you'll get a y factor from one of your limits. You can then do the integration with a simple substitution d(a y^2 + b) = 2y dy.

You'll have another term however with no "extra y". To integrate [itex]\sqrt{a y^2 + b}dy[/itex] you'll need a trig substitution.
First divide out the coefficient:
[itex]\sqrt{a(y^2+b/a)}[/itex]
and then its a matter of integrating [itex] \int \sqrt{y^2 + c^2}dy[/itex].
Try the trig substitution y = c * tan(theta).
(Or if you like hyperbolic trig, y = c * sinh(z).)